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Derivative of Inverse Function

  • #1

Homework Statement



Find g' (-1/2), where g(x) is the inverse of f(x) = x3 / (x2 + 1)

Homework Equations



g'(x) = 1/f' [g(x)]

The Attempt at a Solution



Everything else I tried got really messy, but I feel like I'm going out on a limb by:

1) y=x3 / (x2+1)

2) y=x3x-2 / 1

3) y=x

4) g '(0.5)= 1/ (0.53 /(0.52+1) = 10

I don't even know if I'm going about it the right way. Any help would be awesome. Thanks.
 

Answers and Replies

  • #2
128
0
You say in 1) that
y = x^3 / (x^2 + 1)
and in 3) that
y = x.

If you put in x = 1 you see that in 1) you get 1/2 and in 3) you get 1. Something must hence be wrong.
 
  • #3
Very true. I realize something is wrong. That was just my attempt.
 
  • #4
But . . . I could still use some help with this problem, obviously.
 
  • #5
fzero
Science Advisor
Homework Helper
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1) y=x3 / (x2+1)

2) y=x3x-2 / 1
Line 2 is incorrect, it should be

y=x3x-2 / (1+x-2),

which actually just makes things more complicated. Just differentiate the expression on line 1 directly.
 
  • #6
612
1
It would help to figure out what g(-1/2) was equal to

Then you could just plug stuff into this relationship: g'(x) = 1/f' [g(x)] -> g'(-1/2) = 1/f' [g(-1/2)]

Since we know that g is the inverse of f, we should try to solve when f(a) = -1/2. Then we would know a = g(-1/2).

So -1/2 = x^3 / (x^2 + 1) -> -1/2*(x^2 + 1) = x^3 -> 0 = x^3 + 1/2x^2 + ½ which clearly has a solution at -1. So we know g(-1/2) = -1 which means: g'(-1/2) = 1/f'(-1) and you can do the rest from there id bet.
 
  • #7
So I differentiate to get

(x2+1(3x2))-(2x(x3))/((x2+1)2)

And then I dared to simplify it down to

(x4+3x2)/((x2+1)2)

But what I don't understand from all of this is that if I do that, am I plugging this in for the g(x) for the 1/ f '[g(x)]? That would be like plugging f '[f '(x)] right? I'm sorry I'm just trying to understand how to solve these types of problems.
 
  • #8
128
0
As the guy above said, if f(c) = -1/2 then g(-1/2) = c. To find out what g(-1/2) is we need to find c, that is to solve the equation f(x) = -1/2.

When this is done you can just plug everything into the formula.
 
  • #9
1,796
53
So I differentiate to get

(x2+1(3x2))-(2x(x3))/((x2+1)2)

And then I dared to simplify it down to

(x4+3x2)/((x2+1)2)

But what I don't understand from all of this is that if I do that, am I plugging this in for the g(x) for the 1/ f '[g(x)]? That would be like plugging f '[f '(x)] right? I'm sorry I'm just trying to understand how to solve these types of problems.
Surely if I have a function y=f(x), then the inverse is just x=g(y) and it's derivative is [itex]\frac{dx}{dy}[/itex] which is a function of y. So that for:

[tex]f(x)=\frac{x^3}{x^2+1}[/tex]

we have [itex]x=f^{-1}(y)[/itex] and so the derivative of the inverse at [itex]y_0[/itex] can be written as:

[tex]\frac{df^{-1}}{dy}\biggr|_{y_0}=\frac{dx}{dy}\biggr|_{y_0}=\frac{1}{\frac{dy}{dx}}\biggr|_{x(y_0)}[/tex]

So now let [itex]y_0=-1/2[/tex]. Then solve:

[tex]y_0=-1/2=\frac{x^3}{x^2+1}[/tex]

which has three solutions. The inverse and it's derivative are multi-valued right since it's going to be a cube root correct? Therefore, just find the three roots [itex] (x_1,x_2,x_3)[/itex] and for example, then one derivative of the inverse at y=-1/2 is:

[tex]\frac{df^{-1}}{dy}\biggr|_{y=-1/2}=\frac{dx}{dy}\biggr|_{y=-1/2}=\frac{1}{\frac{dy}{dx}}\biggr|_{x(-1/2)=x_1}[/tex]

Try and follow the following Mathematica code which explicitly computes the inverse, then finds it's (multivalued) derivative at y=-1/2:

Code:
f[x_] := x^3/(x^2 + 1); 
mysols = x /. Solve[f[x] == y, x]
N[D[mysols, y] /. y -> -2^(-1)]
 
Last edited:

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